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Ivenika [448]
3 years ago
14

PLEASEEEE HELPPP!!!!

Mathematics
1 answer:
CaHeK987 [17]3 years ago
5 0

Answer:

The slope of this line is 1/2.

Step-by-step explanation:

You have to pick any two coordinates from the line. So I will pick (2,0) and (4,1). Then you have to use gradient formula :

m =  \frac{y2 - y1}{x2 - x1}

Let (x1,y1) be (2,0),

Let (x2,y2) be (4,1),

m =  \frac{1 - 0}{4 - 2}

m =  \frac{1}{2}

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zzz [600]

I think the answer Is B

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3 years ago
Simplify the expression using the order of operations. 3 (8-4)2 + 7.9
OlgaM077 [116]

Answer:31.9

Step-by-step explanation:

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Given: △ACD, m∠BCA = 60° CB ⊥ AD , m∠BDC = 45° AC = 10 cm. Find: BC and DC
neonofarm [45]

Answer:

See if this will help,

https://www.pmschools.org/site/handlers/filedownload.ashx?moduleinstanceid=701&dataid=3612&FileName=2015%20triangle%20proofs%20answers.pdf



8 0
3 years ago
11) Help pls I have the answer I just need to show the work.
spayn [35]
True.
The Pythagorean theorem is a^2 + b^2 = c^2 where a^2 and b^2 are the two legs of the triangle (two sides connected by the right angle) and c is the hypotenuse (longest side, opposite the right angle).

To solve using the Pythagorean theorem, plug in sides AC and CB into a and b, then solve for c.

23^2 + 31^2 = c^2
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I hope this helps!
4 0
3 years ago
Simplify the expression. Quantity cosecant of x to the power of two times secant of x to the power of two divided by quantity se
Art [367]

Answer:

D. 1

Step-by-step explanation:

We have the expression, \frac{\csc^{2}x\sec^{2}x}{\sec^{2}x+\csc^{2}x}

We get, eliminating the cosecant function,

\frac{\sec^{2}x}{\frac{\sec^{2}x}{\csc^{2}x}+1}

As, sinx is reciprocal of cosecx and cosx is reciprocal of secx,

i.e. \frac{\sec^{2}x}{\frac{\sin^{2}x}{\cos^{2}x}+1}

i.e. \frac{1}{\cos^{2}x}\times \frac{\cos^{2}x}{\sin^{2}x+\cos^{2}x}

Since, we know that, \sin^{2}x+\cos^{2}x=1

Thus,

\frac{1}{\cos^{2}x}\times \frac{\cos^{2}x}{\sin^{2}x+\cos^{2}x}=1

So, after simplifying, we get that the result is 1.

Hence, option D is correct.

3 0
3 years ago
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